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Polar coordinates: derivation from rotation group

  1. Feb 2, 2010 #1
    I posted a similar question long time ago, but after working on it I am still unable to arrive at a solution.
    Let's have a group of linear transformations (rotations in the xy-plane):

    [tex]R_\theta=\{ (\begin{array}{ccc} cos\theta & -sin\theta \\ sin\theta & cos\theta \end{array}) \\ : \\ \theta \in [0,2\pi] \}[/tex]

    The question is: How can I construct an orthogonal curvilinear coordinates system, in which the parameter [itex]\theta[/itex] works as one coordinate?
    What I am supposed to get as a result are essentially the equations defining the cartesian-to-polar transformation.

    My attempt:
    Observe that given any vector x, the orbit [tex]R_{\theta}(\mathbf{x})[/tex] is a parametric curve which is obviously a circle.
    The (gradient) vectors

    [tex]e_\theta=\frac{\partial R_{\theta}(\mathbf{x})}{\partial \theta}[/tex] are tangent to the curve, so if we consider their orthogonal complement [tex]e_\theta^*[/tex] (which is easy to find), we have already found a family of local orthogonal bases.
    How can I continue from this point???
    I am supposed to get: [itex]r = (x^2 + y^2)^{1/2}[/itex] and [itex]\theta = atan2(y/x) [/itex], but I don't know how to arrive at that.
  2. jcsd
  3. Feb 4, 2010 #2
    Since you do not include any transformation in the radial direction, you can only use it to create a coordinate system on the sphere S^1, using your coordinate theta. This is done choosing a point of origin for you coordinate system, e.g. (x,y) = (1,0). From this, just apply a rotation with angle theta, and you get the correspondence between angles theta and the pairs of coordinates (x,y) along the 1-sphere S^1.

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